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# Calorimetry - determining Determining enthalpy and internal energy of formation for K2Opotassium superoxide

Given $$2.14\ \mathrm g$$ of $$\ce{K(s)}$$, determine $$\Delta H^\circ_\mathrm{f,m}$$ and $$\Delta U^\circ_\mathrm{f,m}$$ for $$\ce{K2O}$$.

We know:

• The calorimeter's constant: $$1849\ \mathrm{J\cdot K^{-1}}$$
• The mass of water inside it: $$1450\ \mathrm g$$
• The change in temperature: $$2.62\ \mathrm K$$
• The end product is $$\ce{K2O}$$

We know:

• The calorimeter's constant: $$1849\ \mathrm{J\cdot K^{-1}}$$
• The mass of water inside it: $$1450\ \mathrm g$$
• The change in temperature: $$2.62\ \mathrm K$$
• The end product is $$\ce{K2O}$$

The process should be: determining mol of $$\ce{K(s)}$$, which is $$0.054\ \mathrm{mol}$$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $$2.62\ \mathrm K$$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $$322\ \mathrm{kJ}$$ or $$96.8\ \mathrm{kJ}$$.

Continuing with this reasoning, the $$\Delta U^\circ_\mathrm{f,m}$$ should be $$755\ \mathrm{kJ}$$ or $$193\ \mathrm{kJ}$$, right? What do we need to obtain $$\Delta H^\circ_\mathrm{f,m}$$?

# Calorimetry - determining enthalpy and internal energy of formation for K2O

Given $$2.14\ \mathrm g$$ of $$\ce{K(s)}$$, determine $$\Delta H^\circ_\mathrm{f,m}$$ and $$\Delta U^\circ_\mathrm{f,m}$$ for $$\ce{K2O}$$.

We know:

• The calorimeter's constant: $$1849\ \mathrm{J\cdot K^{-1}}$$
• The mass of water inside it: $$1450\ \mathrm g$$
• The change in temperature: $$2.62\ \mathrm K$$
• The end product is $$\ce{K2O}$$

The process should be: determining mol of $$\ce{K(s)}$$, which is $$0.054\ \mathrm{mol}$$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $$2.62\ \mathrm K$$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $$322\ \mathrm{kJ}$$ or $$96.8\ \mathrm{kJ}$$.

Continuing with this reasoning, the $$\Delta U^\circ_\mathrm{f,m}$$ should be $$755\ \mathrm{kJ}$$ or $$193\ \mathrm{kJ}$$, right? What do we need to obtain $$\Delta H^\circ_\mathrm{f,m}$$?

# Determining enthalpy and internal energy of formation for potassium superoxide

Given $$2.14\ \mathrm g$$ of $$\ce{K(s)}$$, determine $$\Delta H^\circ_\mathrm{f,m}$$ and $$\Delta U^\circ_\mathrm{f,m}$$ for $$\ce{K2O}$$.

We know:

• The calorimeter's constant: $$1849\ \mathrm{J\cdot K^{-1}}$$
• The mass of water inside it: $$1450\ \mathrm g$$
• The change in temperature: $$2.62\ \mathrm K$$
• The end product is $$\ce{K2O}$$

The process should be: determining mol of $$\ce{K(s)}$$, which is $$0.054\ \mathrm{mol}$$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $$2.62\ \mathrm K$$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $$322\ \mathrm{kJ}$$ or $$96.8\ \mathrm{kJ}$$.

Continuing with this reasoning, the $$\Delta U^\circ_\mathrm{f,m}$$ should be $$755\ \mathrm{kJ}$$ or $$193\ \mathrm{kJ}$$, right? What do we need to obtain $$\Delta H^\circ_\mathrm{f,m}$$?

3 typography corrected

Given $$2.14g$$ of $$K_{(s)}$$, determine $$\Delta Hº_{f,m}$$ and $$\Delta Uº_{f,m}$$ for $$K_2O$$.

Given $$2.14\ \mathrm g$$ of $$\ce{K(s)}$$, determine $$\Delta H^\circ_\mathrm{f,m}$$ and $$\Delta U^\circ_\mathrm{f,m}$$ for $$\ce{K2O}$$.

We know:

• The calorimeter's constant: $$1849J•K^{-1}$$$$1849\ \mathrm{J\cdot K^{-1}}$$
• The mass of water inside it: $$1450g$$$$1450\ \mathrm g$$
• The change in temperature: $$2.62K$$$$2.62\ \mathrm K$$
• The end product is $$K_2O$$$$\ce{K2O}$$

The process should be: determining mol of $$K_{(s)}$$$$\ce{K(s)}$$, which is $$0.054mol$$$$0.054\ \mathrm{mol}$$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $$2.62K$$$$2.62\ \mathrm K$$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $$322kJ$$$$322\ \mathrm{kJ}$$ or $$96.8kJ$$$$96.8\ \mathrm{kJ}$$.

Continuing with this reasoning, the $$\Delta Uº_{f,m}$$$$\Delta U^\circ_\mathrm{f,m}$$ should be $$755kJ$$$$755\ \mathrm{kJ}$$ or $$193kJ$$$$193\ \mathrm{kJ}$$, right? What do we need to obtain $$\Delta Hº_{f,m}$$$$\Delta H^\circ_\mathrm{f,m}$$?

Given $$2.14g$$ of $$K_{(s)}$$, determine $$\Delta Hº_{f,m}$$ and $$\Delta Uº_{f,m}$$ for $$K_2O$$.

We know:

• The calorimeter's constant: $$1849J•K^{-1}$$
• The mass of water inside it: $$1450g$$
• The change in temperature: $$2.62K$$
• The end product is $$K_2O$$

The process should be: determining mol of $$K_{(s)}$$, which is $$0.054mol$$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $$2.62K$$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $$322kJ$$ or $$96.8kJ$$.

Continuing with this reasoning, the $$\Delta Uº_{f,m}$$ should be $$755kJ$$ or $$193kJ$$, right? What do we need to obtain $$\Delta Hº_{f,m}$$?

Given $$2.14\ \mathrm g$$ of $$\ce{K(s)}$$, determine $$\Delta H^\circ_\mathrm{f,m}$$ and $$\Delta U^\circ_\mathrm{f,m}$$ for $$\ce{K2O}$$.

We know:

• The calorimeter's constant: $$1849\ \mathrm{J\cdot K^{-1}}$$
• The mass of water inside it: $$1450\ \mathrm g$$
• The change in temperature: $$2.62\ \mathrm K$$
• The end product is $$\ce{K2O}$$

The process should be: determining mol of $$\ce{K(s)}$$, which is $$0.054\ \mathrm{mol}$$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $$2.62\ \mathrm K$$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $$322\ \mathrm{kJ}$$ or $$96.8\ \mathrm{kJ}$$.

Continuing with this reasoning, the $$\Delta U^\circ_\mathrm{f,m}$$ should be $$755\ \mathrm{kJ}$$ or $$193\ \mathrm{kJ}$$, right? What do we need to obtain $$\Delta H^\circ_\mathrm{f,m}$$?

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